Answer
$$\displaystyle{\int_{0}^{\frac{\pi }{2}}\frac{\cos t}{\sqrt{1+\sin^2t}}dt=\ln\left(1+\sqrt{2}\right)}\\$$
Work Step by Step
$\displaystyle{I=\int_{0}^{\frac{\pi }{2}}\frac{\cos t}{\sqrt{1+\sin^2t}}dt}\\$
$\displaystyle \left[\begin{array}{ll} \sin t=\tan\theta & \sin^2t=\tan^2\theta \\ & \\ \frac{dt}{d\theta}\cos t=\sec^2\theta & dt=\frac{\sec^2\theta}{\cos t}\ d\theta \end{array}\right]$ Integration by substitution
$\displaystyle{I=\int_{0}^{\frac{\pi }{4}}\frac{\cos t}{\sqrt{1+\tan^2\theta}}\times\frac{\sec^2\theta}{\cos t}d\theta}\\ \displaystyle{I=\int_{0}^{\frac{\pi }{4}}\frac{\sec^2\theta}{\sec\theta}d\theta}\\ \displaystyle{I=\int_{0}^{\frac{\pi }{4}}{\sec\theta}\ d\theta}\\ \displaystyle{I=\left[\ln\left|\sec\theta+\tan\theta\right|\right]_{0}^{\frac{\pi }{4}}}\\ \displaystyle{I=\ln\left(1+\sqrt{2}\right)-\ln1}\\ \displaystyle{I=\ln\left(1+\sqrt{2}\right)} $